CH301 Fall 2012 Name: KEY VandenBout/LaBrake UNIT 2 READINESS ASSESSMENT QUIZ (RAQ) THIS QUIZ WILL BE PACED WITH CLICKER QUESTIONS

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1 CH301 Fall 2012 Name: KEY VandenBout/LaBrake UNIT 2 READINESS ASSESSMENT QUIZ (RAQ) THIS QUIZ WILL BE PACED WITH CLICKER QUESTIONS 1. A laser pulse shines for 10 s delivering a total energy of 4 mj of 633 nm light. Another laser delivers the same amount of energy with a wavelength of 408 nm. a) Which laser is delivering more photons to the sample? Red laser, 633 nm laser because the total energy delivered for the two lasers in the same. Need more photons of lower energy to get up to 4 mj. b) How much energy per photon (in units of ev) will be delivered for the red laser versus the blue laser? (1eV = x J) E = hc/λ =hc/633x10-9 x x = 1.96 ev E = hc/λ =hc/408x10-9 x x = 3.04 ev c) Each of these lasers shines on calcium (Φ = 2.90 ev). What will happen when a 10 s pulse of red laser shines on calcium? Nothing, photons don t have enough energy to eject the electrons What will happen when a 10 s pulse of the blue laser shines on calcium? Electrons will be ejected with some KE, that can be calculated. 2. Please explain the change in effective nuclear charge, Z eff, as you move across a row in the periodic table from left to right. Indicate how this change in Z eff affects the ionization energy and the atomic radii of the atoms as you move across a row. Use the elements calcium and selenium as specific examples predicting which would have the smaller atomic radius and why. Z eff = nuclear charge inner shell e - From left to right across the periodic table, Z eff INCREASES As we go from left to right, we add protons, yet since we are in the same row, we have not added any inner shell electrons. Therefore, Z eff increases As Z eff increases, there is a stronger pull on the electrons. Ca has a Z eff = 20 ( ) = 2 Se has a Z eff = 34 ( ) = 6 Selenium has a larger Z eff, and is in the same row as Ca. Even though there are more valence electrons in Se, there is a larger Z eff holding them in. This larger Z eff will hold the electrons closer to the nucleus, creating a smaller radius

2 A larger Z eff holds the electrons tighter. The tighter the electrons are held to the atom, the more energy it takes to remove an electron. This is ionization energy! The higher the Z eff, the higher the ionization energy, the smaller the atomic radius. 3. Please write out the electron configuration for Germanium, Ge, using two different methods, the orbital notation (including using dashes and arrows, no noble gas shorthand), and the noble gas short hand method (don t need to use dashes and arrows, just superscript numbers will do). State the number of valence electrons in Ge. Orbital Notation: 1s 2s 2p 3s 3p 4s 3d 4p [Ar] 4s 2 3d 10 4p 2 Valence electrons: 4 (2 from 4s subshell, 2 from 4p subshell) 4. Please draw the Lewis structure for the molecule, oxalate ion, C 2 O State whether or not you expect all the CO bond lengths to be the same or different, and explain your answer. Calculate the formal charge for each atom in the structure. The sum of the formal charges is what? S = N A Needed: 2C x 8 = 16 4O x 8 = 32 Total Needed: = 48 Available: 2C x 4 = 8 4O x 6 = 24 Total Available: = 32 Shared electrons = = 14 14/2 = 7 bonds

3 Formal Charge = Valence Electrons (Lone electrons + ½Shared electrons) OR FC = Group Number (Lone electrons + number of bonds) Lone electrons NOT the number of lone pairs For either carbon: F = 4 (0 + 4) = 0 For double bond oxygen: F = 6 (4 + 2) = 0 For single bond oxygen: F = 6 (6 + 1) = -1 So two of the oxygens have a formal charge of -1, giving the ion a total overall charge of -2 All CO bond lengths should be the same. Each Lewis structure is one resonance structure. The actual structure is an average of all of these. 5. Using a drawing and your own words, articulate to the best of your ability, the difference between an ionic compound and a covalent compound. Na + Cl See the last page for a better picture!

4 In an ionic compound, atoms or polyatomic ions have opposite charges and form a crystal lattice, and the chemical formula in fact is a formula unit, or a ratio of the elements. Remember, when we think about ionic compounds, we are thinking about a solid crystal, not the individual ions (we do not care what happens when salt dissolves in water when describing an ionic bond) The atoms actually have opposite charges. No electrons are shared. See the next page for a better picture! H Cl In a covalent compound, electrons are actually shared between atoms, forming a chemical bond. The sharing may not be equal between the two atoms, based on their electronegativity, but the electrons are always shared between the atoms.

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